# Homework Help: Gauss's Law

1. Jun 12, 2007

### flash

1. The problem statement, all variables and given/known data

Use Gauss' law to find the charge density on a Van Der Graff dome (r=40cm) if it is charged to 100kV. What is the electric field strength at r=25cm?

I understand gauss's law and I know that I need to use it to find the total charge enclosed on the dome. I can then work out the surface area of the dome and calculate charge density.

If I define a gaussian surface to be a sphere outside the dome, I can work out the charge enclosed if i know the electric field at the gaussian surface. I figure I must use the 100kV given to find the field at the gaussian surface, but I don't understand how.

Also, I need a little help with the second part of the problem. That radius given is inside the dome, unsure of where to go there.

Thanks for any help.

2. Jun 12, 2007

### Mindscrape

So, as a starting point, what are the equations that you have to work with?

Also, as a matter of curiosity, what level of physics is this (I'm assuming a university calc based intro E&M), and what math have you taken?

3. Jun 12, 2007

### jackiefrost

Check out http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potsph.html" [Broken]

Last edited by a moderator: May 2, 2017
4. Jun 12, 2007

### nrqed

That's probably a stupid question but I am wondering...is the sphere meant to be hollow or full? If it's hollow (which makes more sense), then the E field inside the dome will be zero since there is no net charge enclosed (I am assuming a complete sphere, in an actual VdG there is of course a part of the sphere cut out but here I assume they mean to ignore that). The 100 kV is not needed. So I must not understand the question!

5. Jun 12, 2007

### siddharth

I think Van der Graaff generators usually use a hollow sphere. In any case, they use a metal sphere, so any charge would be on the outside of the sphere.

6. Jun 12, 2007

### nrqed

Ok. So whether it is a hollow sphere or a full conducting sphere, the E field at 25 cm would be zero with no calculation required. (I guess the 100 kV was just needed to find the total charge)